I got this problem in an exam and it looks so simple. Of course the cauchy’s formula for a holomorphic function $f$ allows one to say that $|f|$ is sub harmonic. But I am not able to see why it should be strictly sub-harmonic.
Any help or hint would be appreciated.
Let $\overline {B}(z,r) $ be contained in the domain of $f$. Theb $f(z+re^{i\theta}) =\sum a_n(re^{i\theta})^{n}$ where $a_n=\frac {f^{(n)}(z)} {n!}$. Integrating w.r.t. $\theta$ from $0$ to $\pi$ we get $f(z)=\frac 1 {2\pi}\int_0^{2\pi} f(z+re^{i\theta})d\theta$. Hence $|f(z)|\leq \frac 1 {2\pi}\int_0^{2\pi} |f(z+re^{i\theta}|d\theta$. This proves that $|f|$ is subharmonic. Now suppose equality holds. Then $f(z+re^{i\theta})$ is real (and has the same sign) for all $\theta$. Apply MMP to $e^{if}$ and $e^{-if}$ to conclude that $f$ is a constant on $B(z,r)$ (and hence on the entire domain of $f$ as log as the domain is connected). This proves that $|f|$ is strictly subharmonic unless $f$ is a constant.
Of course constants are holomorphic but $|f|$ is not strictly subharmonic when $f$ is a constant.